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Motion of Connected Bodies

Question 1:
Two blocks are in contact on a frictionless table. One has mass m and the other 2 m. A force F is applied on 2 m as shown in the figure. Now the same force F is applied from the right on m. In the two cases respectively, the ratio of force of contact between the two blocks will be:

1:1

1:2

2:1

1:3

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Solution:

Acceleration of the blocks in both the case will be F/3m.

N1= m.a

N2= 2m.a

so N1/N2= 1/2

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Question 2:
Two forces of 6N and 3N are acting on the two blocks of 2kg and 1 kg kept on frictionless floor. What is the force exerted on 2kg block by 1 kg block?

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Solution:

Net force is 6N-3N= 3N

Accleration of the system is 3N/3Kg = 1 m/s^2

Net Force on 1kg object is N1 - 3= 1*1 so, N1 = 4N

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Question 3:
A body of mass 8 kg is hanging from another body of mass 12 kg. The combination is being pulled by a string with an acceleration of 2.2 ms^-2. The tension T₁ and T₂ will be respectively: (use g = 9.8m/s²)

200 N, 80 N

220 N, 90 N

240 N, 96 N

260 N, 96 N

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Solution:

Taking both the objects as one system

T1- 20g=20a ⇒ T1= 20(g+a)=20*12= 240 N

Now taking lower object as a different system

T2-8g=8a ⇒T2=8(g+a)=8*12= 96N

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Question 4:
Two blocks of masses 1 kg and 2 kg are connected with massless spring as shown. If the acceleration of 1 kg block is 1 m/s² towards right then the acceleration of 2 kg block is:

2 m/s²

1 m/s²

3 m/s²

5 m/s²

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Solution:

Spring force acting on 1 kg block is F = m.a = 1 N

NLM Question NEET Physics

Same spring force will act on 2 kg block in opposite direction. Thus net force acting on 2 kg block is 5N - 1 N = 4N

Acceleration of the block will be 4 N/2 kg = 2 m/s²

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Question 5:
Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is
applied on the 4 kg block, then the contact force between A and B is

18 N

2 N

8 N

6 N

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Solution:

The total mass of the system:

$M = 4 + 2 + 1 = 7 \text{ kg}$

The acceleration of the system:

$a = \frac{\text{Total Force}}{\text{Total Mass}} = \frac{14}{7} = 2 \text{ m/s}^2$

Now, considering block B and C together (mass = $2 + 1 = 3$ kg), the force required to accelerate them:

$F_{AB} = (3 \times 2) = 6 \text{ N}$

Thus, the contact force between A and B is 6 N.

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